CHAPTER 2: OPTIMAL PIPE DIAMETER FOR A
SINGLE PIPE SEGMENT
To find the optimal diameter for a single pair of supply and return pipes, we need
to consider the costs involved and minimize their sum with respect to the pipe
diameter. The cost minimization is done for the life cycle of the system using a net
present value approach. Some types of heat distribution systems may have a salvage
value, while others will, in fact, have a disposal cost associated with the end of their
useful lifetime. Since these will in general be mild functions of the pipe diameter,
they will not significantly affect the optimal pipe diameter and thus will not be
treated here. With these limitations in mind, our objective function, the total life
cycle cost, becomes
min. Ct = Chl + Cpe + Cpp + Cm&r
(2-1)
where Ct = total system owning and operating cost ($)
Chl = cost of heat losses ($)
Cpe = cost of pumping energy ($)
Cpp = capital costs of pipes and pumps ($)
Cm&r = cost of maintenance and repair ($).
Now let's look at each of the costs in eq 2-1 in detail, starting with the cost of heat
losses.
COST OF HEAT LOSS
The basic form of the heat loss cost is
Chl = PVFh ∫yr ChQhldt
(2-2)
where Ch
= cost of heat ($/Wh)
PVFh
= present value factor for heat (dimensionless)
Qhl
= rate of heat loss (W)
= time of year (hr [0 ≤ t ≤ 8760]).
t
In the most general case, the cost of heat Ch can be a function of time because of
seasonal usage rates. The rate of heat loss Qhl will also be a function of time over the
heat losses as the system ages. This can not be incorporated directly into the
formulation as given above, but could be allowed for by using an appropriate
escalation factor in the present value factor for heat costs PVFh.
The only variable defined above that is dependent on our decision variable, the
pipe diameter d, is the heat loss rate itself Qhl. For a single buried pipe, the
relationship is
Qhl = L(Tp Tg)/Ro
(2-3)
where Tp = pipe outer surface temperature (C)
Tg = soil temperature (C)
L = pipe length (m).
5
Previous Page
